john4 wrote:

If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

A. 25

B. 27

C. 45

D. 75

E. 125

Given \(n(2^5)(6^2)(7^3)\) is divisible by both \(5^2\) and \(3^3\).

Therefore \(n(2^5)(6^2)(7^3)\) will be divisible by LCM of \(5^2\) and \(3^3\).

LCM of \(5^2\) and \(3^3 = (5^2)(3^3)\)

By Simplifying the expression we get;

\(\frac{n(2^5)(6^2)(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^5)(2*3)^2(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^5)(2^2)(3^2)(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^7)(3^2)(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^7)(7^3)}{(5^2)(3)}\)

\(2\) and \(7\) are not divisible by \(5\) or \(3\). Hence \(n\) should be a multiple divisible by \((5^2*3)\)

Therefor \(n(2^7)(7^3)\) to be divisible by \((5^2*3)\), smallest possible of \(n\) Should be \(= 5^2*3 = 25*3 = 75\)

Answer (D)..._________________

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